Difference between revisions of "2000 AMC 12 Problems/Problem 17"
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Revision as of 15:37, 14 June 2013
Problem
A circle centered at has radius and contains the point . The segment is tangent to the circle at and . If point lies on and bisects , then
Solution
Since is tangent to the circle, is a right triangle. This means that , and . By the Angle Bisector Theorem, We multiply both sides by to simplify the trigonometric functions, Since , . Therefore, the answer is .
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |